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Magazine Chapter 6: Problem 23
The convolution product can be defined for other classes of functions. Thus \(f \times g\) exists as an element in \(C_{0}(G)\) whenever \(f \in \mathscr{L}^{p}(G)\) and \(g{g} \in \mathscr{L}^{q}(G)\) with \(p^{-1}+q^{-1}=1\) [because \(f \times g(x)=\int f(y)_{x} \check{g}(y) d y\) ]. In the case \(p=1\), \(q=\infty\), however, we only have \(f \times g\) as a uniformly continuous, bounded function on \(G\). We wish to define the convolution product of finite Radon charges (cf. 6.5.8). If \(\Phi, \Psi \in M(G)\), we define \(\Phi \otimes \Psi\) as a finite Radon charge on \(G \times G\), either by mimicking the proof of \(6.6 .3\) or by taking polar decompositions \(\Phi=|\Phi|\left(u^{*}\right)\) and \(\Psi=|\Psi|\left(v^{\circ}\right)\) as in \(6.5 .6\) and \(6.5 .8\) and then setting \((\Phi \otimes \Psi) h=(|\Phi| \otimes|\Psi|)((u \otimes v) h), \quad h \in C_{c}(G \times G) .\) \((\Phi \times \Psi) f=(\Phi \otimes \Psi)(f \circ \pi), \quad f \in C_{c}(G)\), \((\Phi \times \Psi) f=(\Phi \otimes \Psi)(f \circ \pi), \quad f \in C_{c}(G)\), \(\rightarrow G\) is the product map \(\pi(x, y)=x y .\) Note that although where \(\pi: G \times G \rightarrow G\) is the product map \(\pi(x, y)=x y .\) Note that although \(f \circ \pi\) is a bounded continuous function on \(G \times G\) it does not belong to \(C_{c}(G \times G)\) (if \(f \neq 0\) ), so that we need the assumption that \(\Phi\) and \(\Psi\) are finite Having done this, we define the product in \(M\) $$ (\Phi \times \Psi) f=(\Phi \otimes \Psi)(f \circ \pi) \text {, } $$ where \(\pi: G \times G \rightarrow G\) is the product map \(\pi(x, y)\) \(f \circ \pi\) is a bounded continuous function on \(G \times\) \(C_{c}(G \times G)(\) if \(f \neq 0)\), so that we need the assumption that charges. It follows from \((*)\) and (**) that \(\Phi \times \Psi \in M(G)\), with $$ \|\Phi \times \Psi\| \leq\|\Phi \otimes \Psi\|=\|\Phi\|\|\Psi\| \text {. } $$ $$ ((\Phi \times \Psi) \times \Omega) f=(\Phi \otimes \Psi \otimes \Omega)(f \circ \tau)=(\Phi \times(\Psi \times \Omega)) f $$ where \(\tau(x, y, z)=x y z\), so that the product is associative. In conjunction with \(6.5 .9\) this shows that \(M(G)\) is a unital Banach algebra (the point measure \(\delta_{1}\) at 1 being the unit). Defining $$ \Phi^{*} f=\overline{\Phi f}, \quad f \in C_{c}(G), $$ where \(f(x)=\overline{f\left(x^{-1}\right)}\), we see that $$ \left(\Psi * \otimes \Phi^{*}\right)(f \otimes g)=\overline{\Psi(f) \Phi(\tilde{g})}=\overline{(\Phi \otimes \Psi)(\tilde{g} \otimes f)}=\overline{(\Phi \otimes \Psi)(f \otimes g)^{s}} $$ where \(h^{s}(x, y)=h\left(y^{-1}, x^{-1}\right)\) for every function on \(G \times G\). It follows from (**) that for each \(f\) in \(C_{c}(G)\) we have $$ \begin{aligned} \left(\Psi^{*} \times \Phi^{*}\right) f &=\left(\Psi^{\otimes} \otimes^{*}\right)(f \circ \pi)=\overline{(\Phi \otimes \Psi)(f \circ \pi)^{s}} \\ &=\overline{(\Phi \otimes \Psi)(f \circ \pi)}=(\Phi \times \Psi)^{*} f \end{aligned} $$ so that \(*\) is an isometric involution on \(M(G)\).
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